Center of circle is ( 1, 0).
Slope of normal passing though ( 1, 0 ) and ( 2, -5 ) is :
[tex]m_p=\dfrac{0-(-5)}{1-2}\\\\m_p=-5[/tex]
So, slope of tangent will be :
[tex]m_t=-\dfrac{1}{m_n}\\\\m_t=\dfrac{1}{5}[/tex]
Equation of tangent :
[tex]y-(-5)=\dfrac{1}{5}(x-2)\\\\5y+25=x-2\\\\5y-x+27=0[/tex]
Hence, this is the required solution.
